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Question Number 186253 by ajfour last updated on 02/Feb/23

Commented by ajfour last updated on 02/Feb/23

$T h e c u r v e i s a c u b i c (y = x^{3} - x),$ $a n d t h a t 0 < c < \frac{2}{3 \sqrt{3}} \cdot$ $L e t c i r c l e n o t n e c e s s a r i l y b e$ $t a n g e n t t o c u b i c c u r v e a t (q, c) .$
	Answered by a.lgnaoui last updated on 04/Feb/23

	$e q u a t i o n d e d r o i t e (D)$ $y = a x + b x = 1 y = 0$ $E x t r e m u m s = (- \frac{\sqrt{3}}{3}, + \frac{\sqrt{3}}{3})$ $c o n c i d e r o n s B (- \frac{\sqrt{3}}{3}, \frac{2 \sqrt{3}}{9}) \in (D)$ $y = \frac{1 - \sqrt{3}}{3} x + \frac{\sqrt{3} - 1}{3}$ $c e r c l e (C) c o u p e l a c i u r b e e n C$ $C (- q, q - q^{3}), t a n g e n t e e n C est$ $\frac{dy}{dx} (x = - q) = {3 q}^{2} - 1 = \tan θ$ $r + r \cos θ = h - (- q) = h + q$ $r = \frac{h + q}{1 + \frac{1}{\sqrt{1 + \tan^{2} θ}}} = \frac{(h + q) \sqrt{1 + \tan^{2} θ}}{1 + \sqrt{1 + \tan^{2} θ}}$ $r = \frac{(h + q) \sqrt{{(q^{2} - \frac{1}{3})}^{2} + \frac{1}{9}}}{\sqrt{{(q^{2} - \frac{1}{3})}^{2} + \frac{1}{9}} + \frac{1}{3}}$ $l a d r o i t e (D) t a n g e n t e a u c e r c l e (C) e n C$ $\tan ψ = \frac{1 - \sqrt{3}}{3}$ ${(x - x_{0})}^{2} + {(y - y_{0})}^{2} = r^{2}$ $y = \sqrt{r^{2} - {(x - x_{0})}^{2}} + y_{0}$ $\frac{d y}{d x} = \frac{- 2 (x - x_{0})}{2 \sqrt{r^{2} - {(x - x_{0})}^{2}} + y_{0}} = \frac{1 - \sqrt{3}}{3} (1)$ $x - x_{0} = z$ $- 6 z = (\frac{1 - \sqrt{3}}{3}) (y_{0} + 2 \sqrt{r^{2} - z^{2}})$ $\sqrt{r^{2} - z^{2}} = \frac{- 9 z}{1 - \sqrt{3}} - \frac{y_{0}}{2}$ $r^{2} - z^{2} = \frac{81 z^{2}}{{(1 - \sqrt{3})}^{2}} - \frac{9 y_{0}}{(85 - 2 \sqrt{3}) (1 - \sqrt{3})} z + \frac{y_{0}^{2}}{4}$ $[\frac{85 - 2 \sqrt{3}}{(1 - \sqrt{{3)}^{2}}}] z^{2} - \frac{9 y_{0}}{(1 - \sqrt{3)} (85 - 2 \sqrt{3)}} z + \frac{y_{0}^{2}}{4} - r^{2} = 0$ $△ = \frac{81 y_{0}^{2}}{{(1 - \sqrt{3})}^{2} {(85 - 2 \sqrt{3})}^{2}} - \frac{y_{0}^{2} - 4 r^{2}}{{(1 - \sqrt{3})}^{2}} (85 - 2 \sqrt{3})$ $= 81 y_{0}^{2} - (y_{0}^{2} - 4 r^{2}) {(85 - 2 \sqrt{3})}^{3} = 0$ $[81 - {(85 - 2 \sqrt{3})}^{3}] y_{0^{}}^{2} =$ $4 r^{2} {(85 - 2 \sqrt{3})}^{3}$ $y_{0} = \frac{2 r \sqrt{{(85 - 2 \sqrt{3})}^{3}}}{{(85 - 2 \sqrt{3})}^{3} - 81}$ $z = \frac{9 - 9 \sqrt{3}}{2 {(85 - 2 \sqrt{3})}^{2}} y_{0}$ $e n r e m p l a c a n t d a n s z$ $e n d e d u i r e x - x_{0}$ $d a n s l e q u a t i o n d e c e r x l e$ $y = y_{0} + \sqrt{r^{2} - z^{2}}$ $(1) \Rightarrow \frac{1 - \sqrt{3}}{3} = \frac{2 r \sqrt{{(85 - 2 \sqrt{3})}^{3}}}{{(85 - 2 \sqrt{3})}^{3} - 81} +$ $r \sqrt{1 - \frac{81 {(1 - \sqrt{3})}^{2}}{2 (85 - 2 \sqrt{3}) [{(85 - 2 \sqrt{3})}^{3} - 81]}}$ $(A s u i v r e) . . . . . . . . . . . . . . . .$ $a f t e r c a l c u l$ $\frac{1 - \sqrt{3}}{3} ≅ r = \frac{(h + q) \sqrt{{(q^{2} - \frac{1}{3})}^{2} + \frac{1}{9}}}{\sqrt{{(q^{2} - \frac{1}{3})}^{2} + \frac{1}{9}} + \frac{1}{3}}$ $(h + q) \sqrt{(q^{2} - \frac{1}{3}) + \frac{1}{9}} = (\frac{1 - \sqrt{3}}{3}) [\sqrt{{(q^{2} - \frac{1}{3})}^{2} + \frac{1}{9}} + \frac{1}{3}]$ $h + q = (\frac{1 - \sqrt{3}}{3}) \frac{\sqrt{{(q^{2} - \frac{1}{3})}^{2} + \frac{1}{9}} + \frac{1}{3}}{\sqrt{{(q^{2} - \frac{1}{3})}^{2} + \frac{1}{9}}}$ $h = (\frac{1 - \sqrt{3}}{3}) \frac{\sqrt{{(q^{2} - \frac{1}{3})}^{2} + \frac{1}{9}} + \frac{1}{3}}{\sqrt{{(q^{2} - \frac{1}{3})}^{2} + \frac{1}{9}}} - q$ $h = \frac{1 - \sqrt{3}}{3} (1 + \frac{1}{3 \times \sqrt{{(q^{2} - \frac{1}{3})}^{2} + \frac{1}{9}}}) - q$ $h = \frac{1 - \sqrt{3}}{3} + \frac{1 - \sqrt{3}}{\sqrt{{(9 q^{2} - 3)}^{2} + 9}} - q$ $∣ h + q ∣= \frac{\sqrt{3} - 1}{3} + \frac{\sqrt{3} - 1}{\sqrt{{(9 q^{2} - 3)}^{2} + 9}}$ $h + q = r + \frac{r}{\sqrt{1 + \tan θ^{2}}} = r (1 + \frac{1}{\sqrt{1 + {(\frac{1 - \sqrt{3}}{3})}^{2}}})$ $h + q = r + \frac{3 r}{\sqrt{13 - 2 \sqrt{3}}}$ $\frac{\sqrt{3} - 1}{\sqrt{13 - 2 \sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{{(9 q^{2} - 3)}^{2} + 9}}$ ${(9 q^{2} - 3)}^{2} = 4 - 2 \sqrt{3}$ $q^{2} = \frac{\sqrt{4 - 2 \sqrt{3}} + 3}{9}$ $q = \frac{\sqrt{\sqrt{4 - 2 \sqrt{3}} + 3}}{3} = 1, 244$ $d o n c$ $∣ h ∣= 1, 18$ $r = \frac{\sqrt{3} - 1}{3} = 0, 244$
	Commented by ajfour last updated on 04/Feb/23
	$Center of circle Q. Q(q−rsin θ, c+rcos θ) tan θ=3q^2 −1 Tangent: y=−m(x−1) m(1−x)=x^3 −x x^3 +(m−1)x−m=0 double root ⇒ (m^2 /4)=(((1−m)^3 )/(27)) required m=(1/4) Tangent: 4y=1−x h−r=q−rsin θ −(√(17))r=4(c+rcos θ)−((√(1+tan θ))/( (√3)))−rsin θ−1 ((tan θ+1)/3)={4c−1+r(4cos θ−sin θ+(√(17)))}^2 q^3 =q+c$
	$C e n t e r o f c i r c l e Q .$ $Q (q - r \sin θ, c + r \cos θ)$ $\tan θ = 3 q^{2} - 1$ $T a n g e n t : y = - m (x - 1)$ $m (1 - x) = x^{3} - x$ $x^{3} + (m - 1) x - m = 0$ $d o u b l e r o o t \Rightarrow \frac{m^{2}}{4} = \frac{{(1 - m)}^{3}}{27}$ $r e q u i r e d m = \frac{1}{4}$ $T a n g e n t : 4 y = 1 - x$ $h - r = q - r \sin θ$ $- \sqrt{17} r = 4 (c + r \cos θ) - \frac{\sqrt{1 + \tan θ}}{\sqrt{3}} - r \sin θ - 1$ $\frac{\tan θ + 1}{3} = {4 c - 1 + r (4 \cos θ - \sin θ + \sqrt{17})}^{2}$ $q^{3} = q + c$
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